Two circles with center O and P intersects each other in point C and D. Chord AB of the circle with center O touches the circle with center P in point E. Prove that: ∠ADE + ∠BCE = 180°
In the adjoining figure, AD is the bisector of the exterior angle A of triangle ABC. Seg AD intersects the side BC produced in D. Prove that: BD/CD=AB/AC.
Comments
Post a Comment